≈ 2.87
)n
)3 = 150000(1 + 0.1)3 = 150000(1.1)3 = 199650
+ (x - 0)(x - 1)(x - 2)(x - 3)
= 1/3
) has 
exist
≠ 0
does not exist
)
= 
(0/0 form) 
= 
, does not exist
be i.i.d.(independent and identically distributed) random variables with uniform distribution on the interval [0, 1]. Let 
denote the kth order statistic based on the sample 
. If we select two different samples of size 25 and 26, denoted as
and 
respectively, and let Y be the 8th order statistic in both samples. What is the probability that 
?
We label by ai the probability that, starting in state i, the chain is eventually absorbed in the closed set {0,3}. Since {0,3} is closed,
-
a₀ = 1 (if you start in 0 you’re already in {0,3}),
-
a₃ = 1,
and the other closed state 1 lies outside our target class, so -
a₁ = 0.
States 2 and 4 are transient, and satisfy the usual linear equations
a₂ = P(2→0)·a₀ + P(2→1)·a₁
a₄ = Σₖ P(4→k)·aₖ.
From the matrix row for state 2, P(2→0)=1/3, P(2→1)=2/3, and no chance to go directly to 3 or 4. Hence
a₂ = (1/3)·1 + (2/3)·0 = 1/3.
From the row for state 4, the transitions are
-
to 0 with probability 1/8,
-
to 1 with probability 1/8,
-
to 2 with probability 1/2,
-
to 3 with probability 1/8,
-
to 4 itself with probability 1/8.
Thus
a₄ = (1/8)·a₀ + (1/8)·a₁ + (1/2)·a₂ + (1/8)·a₃ + (1/8)·a₄
Substitute a₀=1, a₁=0, a₂=1/3, a₃=1:
a₄ = 1/8 + 0 + (1/2)·(1/3) + 1/8 + (1/8)·a₄
a₄ = 1/8 + 1/6 + 1/8 + (1/8)a₄
Combine the constants: 1/8+1/8 =1/4, and 1/4+1/6 = (3/12+2/12)=5/12, so
a₄ = 5/12 + (1/8)·a₄
Bring the a₄–term to the left:
a₄·(1 – 1/8) = 5/12
(7/8)·a₄ = 5/12
a₄ = (5/12)·(8/7) = 40/84 = 10/21.
Hence starting from state 4 the probability of eventual absorption in {0,3} is 10/21.
is bounded, then f is constant.
and the product 
are bounded, then f is constant.
and 
imposes significant restrictions on the behavior of f. If both these quantities are bounded, then f must be constant.
is bounded, then f is constant.
does not necessarily imply that f is constant. The function f could still vary without violating this condition.
and consider the symmetric bilinear form on R4 given by (v, w) = vt Aw, for v, w ∈ ℝ4. Which of the following statements is true?
, the expression 
follows a chi-squared distribution with 2 degrees of freedom, scaled by σ2 :
where s2 is the sample variance.
and 
are the critical values from the chi-squared distribution with n-1 degrees of freedom.
, where 
denotes a normal distribution with mean μ and variance σ2 .
, where this expression helps define a confidence interval.
, their squares follow a chi-squared distribution with 2 degrees of freedom, scaled by σ2 . Specifically 
can be used to construct a confidence interval for the variance σ2.
is related to the chi-squared distribution. The confidence interval is based on the quantiles of the chi-squared distribution. For a 95% confidence interval, the confidence coefficient is 0.95, which means we need the quantiles of the chi-squared distribution at 5% significance level (0.05).
.
= e
denotes the sample mean and let t0.975, n−1 denote the 0.975-quantile of a Student's− t distribution with n − 1 degrees of freedom. Which of the following statements is true for the following interval 
?
.
is E(Y) = 
and covariance matrix of Y is Σ = 
= 
.......(v)
then the series
is
then
, where r is the perpendicular distance from the z -axis, dm = ρdV is the mass element and ρ is the constant density. For this particular geometry, we are revolving the disc described by the equation 
around the z - axis, where 
.
, a space is compact if every open cover has a finite subcover. Since the cofinite topology ensures that every cover must have at least one large set with a finite complement, X is compact.
with the Euclidean topology is not compact (since 
is not bounded or closed in the Euclidean topology).
and X(n) = max {X1....., Xn}. Which of the following statements are true?
, where X(n) is the maximum value in the sample.
equal to the theoretical mean, we solve for θ.
, where 
is the sample mean.
is the sample mean.
and taking the maximum with -X(1) could ensure that the parameter is positive.
distribution is:
, we get,
, not 
.
, For n ≥ 1, let pn(x) = 1 + x + ....+ xn, Then which of the following statements are true?
, which means pn(x) is an approximation of f(x) as n → ∞.
blows up as x → 1, meaning that as we get closer to 1, the function value grows without bound. This implies that f(x) is not uniformly continuous because the difference in function values for close inputs near 1 can be arbitrarily large.
. Therefore, pn(x) converges to f(x) pointwise on [0, 1).
is continuous and well-behaved on the interval [0, c], and the sequence pn(x) will converge uniformly on compact subintervals away from 1.
; are observed sample means. Under squared error loss function, which of the following statements are true?
